Question

The lowest oxidation state shown by an element in its compound is?
(A) $ -5. $
(B) $ -3. $
(C) $ -4. $
(D) $ -2. $

Answer 1

Hint :We know that the oxidation state of an atom is the number of electrons lost or gained by the atom reacting with any species. Lowest oxidation state corresponds to the maximum number of electrons an atom can gain or lose to form a stable compound with reacting species.

Complete Step By Step Answer:
Electrons that are present in the outermost shell are generally known as valence electrons and the number of valence electrons determines the oxidation state of an atom. We can also say that it is the number of electrons gained or lost from an atom during a chemical reaction. The oxidation state of the elements belonging to the s-block and the p-block of the periodic table are generally calculated as the number of outermost electrons or eight minus the number of outermost electrons. For the d-block and f-block elements, it is determined not only on the basis of outermost electrons but also on d and f orbital electrons.
The state of an element or ion in a compound with regard to the electrons gained or lost by the element or ion in the reaction that formed the compound, expressed as a positive or negative number indicating the charge of the ion is defined as the oxidation state. The lowest oxidation state is $ -4, $ as for carbon in methane or for chromium in $ {{[Cr{{(CO)}_{4}}]}^{4-}} $ where electronic configuration of chromium is given by $ 1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{4}}4{{s}^{2}}. $
Therefore, correct answer is option C i.e. the lowest oxidation state shown by an element in its compound is $ -4. $

Note :
Remember that each transition element can show a minimum oxidation state corresponding to the number of s-electrons and maximum oxidation state equal to the total number of electrons present in both s and d-orbitals. In between oxidation states also become possible such as for Vanadium, $ +2,+3,+4 $ also possible other than $ +5. $